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\title{\vspace{-4cm}\textbf{河北师范大学数学分析真题}}
\author{宁鑫雨}
\date{\today}
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\begin{document}
\date{}
\section*{2015年数学分析}
\begin{problem}[本题30分,每题10分]\\
    1)求极限$\displaystyle\lim_{x\to 0}\left\{\lim_{n\to\infty}\left[\cos\frac{x}{2}\cos\frac{x}{4}\cdots\cdots\cos\frac{x}{2^n}\right]\right\}$\\
2)设$f(x)=
\left\{
\begin{array}{cc}
    x e^{x^{2}},&-\frac{1}{2}<x<\frac{1}{2};\\ 
    -1,&x\leq-\frac{1}{2}\\ 
\end{array}
\right.
$.
求$\displaystyle\int_{\frac{1}{2}}^{2}f(x-\frac{3}{2})\d x$\\
3)计算二重积分$\iint\limits_{D}|y-x|d x d y$,其中$D$为区域$|x|\leq 1$,$0\leq y \leq 2$.\\
    \end{problem}
\begin{solution}
    1)$\displaystyle\Big(\cos\frac{x}{2}\cos\frac{x}{4}\cdots\cos\frac{x}{2^{n}}\Big)\frac{sin\frac{x}{2^{n}}}{sin\frac{x}{2^{n}}}=\frac{sin x}{2^{n}sin\frac{x}{2^{n}}}$\\
    $\therefore$ 原式$=\displaystyle\lim_{x\to0}\left\{\displaystyle\lim_{n\to\infty}\frac{sin x}{2^{n}sin\frac{x}{2^{n}}}\right\}=\lim_{n\to\infty}1=1$\\
    2)$\displaystyle\int_{\frac{1}{2}}^{1}f(x-\frac{3}{2})d x=\int_{\frac{1}{2}}^{2}f(x-\frac{3}{2})d(x-\frac{3}{2})$\\
    令$t=x-\frac{3}{2}$,\\
    $
    \begin{aligned}
        \therefore 
        原式& =\int_{-1}^{\frac{1}{2}}f(t)d t \\
        &=\int_{-1}^{-\frac{1}{2}}f(t)d t+\int_{-\frac{1}{2}}^{\frac{1}{2}}f(t)d t\\
        &=\int_{-1}^{-\frac{1}{2}}-1d t+\int_{-\frac{1}{2}}^{\frac{1}{2}}t e^{t^{2}}d t\\
        &=-\frac{1}{2}+0\\
        &=-\frac{1}{2}
    \end{aligned}$\\
    3)\\
    $ \begin{aligned}
        \iint\limits_{D}|y-x|\d x \d y
                &=\iint\limits_{D_1}(x-y)\d x d y+\iint\limits_{D_2}(y-x)\d x \d y \\
                &=\iint\limits_{D_1}(x-y)\d x d y+\iint\limits_{D_2}(y-x)\d x \d y+\iint\limits_{D_3}(y-x)\d x \d y\\
                &=\int_{0}^{1}\d x\int_{0}^{x}(x-y)\d y+\int_{-1}^{0}\d x\int_{0}^{2}(y-x)\d y+\int_{0}^{1}\d x\int_{x}^{2}(y-x)\d y\\
                &=\frac{1}{6}+3+(3+\frac{1}{6})\\
                &=\frac{19}{3}
        \end{aligned} $
\end{solution}


\begin{problem}[本题10分]
    设$\displaystyle\lim_{n\to\infty}a_{n}=0$,$\displaystyle x_{n}=\frac{2a_1+2^2a_2+\cdots+2^na_n}{2+2^{2}+\cdots+2^{n}}$,
用$\varepsilon-N$定义证明$\displaystyle\lim_{x\to\infty}x_{n}=0$\\
    \end{problem}
\begin{solution}
    证明:\\
    $\because$ $\displaystyle\lim_{n\to\infty}a_{n}=0$\\
    $\therefore$ 任意$\varepsilon>0$,存在自然数$p$,当$n>p$时，$\vert a_{n}\vert<\frac{\varepsilon}{2}$\\
    $\begin{aligned}
        |x_n|=
        &\left|\frac{2a_{1}+2^{2}a_{2}+\cdots+2^{n}a_{n}}{2+2^{2}+\cdots+2^{n}}\right| \\
        &=\left|\frac{2a_{1}+2^{2}a_{2}+\cdots+2^{p}a_{p}}{2+2^{2}+\cdots+2^{n}}+\frac{2^{p+1}a_{p+1}+\cdots+2^{n}a_{n}}{2+2^{2}+\cdots+2^{n}}\right| \\
        &=\left|S_{1}(n)+S_{2}(n)\right|\le\left|S_{1}(n)\right|+\left|S_{2}(n)\right| 
    \end{aligned}$\\
    当$n>p$时,$\displaystyle |S_{2}|<\frac{\varepsilon}{2}\cdot\frac{2^{p+1}+\cdots+2^{n}}{2+2^{2}+\cdots+2^{n}}<\frac{\varepsilon}{2}$\\
    又$\displaystyle\lim_{n\rightarrow\infty}S_{1}(n)=\lim_{n\rightarrow\infty}\frac{2a_{1}+2^{2} a_{2}+\cdots+2^{p}a_{p}}{2+2^{2}+\cdots+2^{n}}=0$\\
    当对上述$\varepsilon>0$,存在$N$,当$n>N$时,有$\left|S_1(n)\right|<\frac{\varepsilon}{2}$\\
    $\therefore$对上述$\varepsilon>0$,存在$N=\min\left\{N_{1},P\right\}$,当$n>N$时\\
    $\displaystyle |x_n|<\vert S_{1}(n)\vert+\vert S_{2}(n)\vert<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$\\
    $\therefore$ $\displaystyle\lim_{x\to\infty}x_{n}=0$
\end{solution}


\begin{problem}[本题10分]
    设$f(x),f^{\prime}(x)$在$[a,b]$上连续,$f^{\prime\prime}(x)$在$(a,b)$内存在，
且$f(a)=f(b)=0$,在$(a,b)$内存在$c\in(a,b)$,使得$f(c)<0$,
证明：在$(a,b)$内存在$\xi$,使得$f^{\prime\prime}(\xi)>0$
    \end{problem}
\begin{solution}
证明：\\
$\because$$f(a)=0,f(c)<0,f(b)=0$\\
$\therefore$由拉格朗日中值定理$\displaystyle f^{\prime}(\eta_{1})=\frac{f(c)-f(a)}{c-a}<0,\eta_{1} \in (a,c)$\\
$\displaystyle f^{\prime}(\eta_{2})=\frac{f(b)-f(c)}{b-c}>0,\eta_{2} \in (c,b)$\\
又$\because$$f^{\prime}(x)$在$[a,b]$上连续,且可导\\
$\therefore$ $\exists\xi\in(\eta_{1},\eta_{2})\subset(a,b),\displaystyle f^{\prime\prime}(\xi)=\frac{f^{\prime}(\eta_{2})-f^{\prime}(\eta_{1})}{\eta_{2}-\eta_{1}}>0$\\
$\therefore$ 原题得证
\end{solution}

    
\begin{problem}[本题10分]
    设$f(x)$在$[0,1]$连续,在$(0,1)$内可导,且满足$\displaystyle f(0)=2015\int_{0}^{\frac{1}{2015}}{}e^{-x^{2}}f(x)d x$\\
    证明：至少存在一点$\xi \in (0,1)$,使得$f^{\prime}(\xi)=2\xi f(\xi)$\\
\end{problem}
\begin{solution}
    证明：\\
  令$F(x)=e^{-x^{2}}f(x)$,则$F(0)=e^{0}f(0)=f(0) $\\
  $\because$$F(x)$在$[0,\frac{1}{2015}]$上连续\\
  $\therefore$由积分第一中值定理，至少存在一点$\eta \in [0,\frac{1}{2015}]$,\\
  使得$\int_{0}^{\frac{1}{2015}}F(x)d x=F(\eta)\frac{1}{2015}$\\
  即有$F(\eta)=2015\int_{0}^{\frac{1}{2015}}F(x)d x=f(0)=F (0)$\\
  又$\because$$F(x)$在$[0,\eta]$上连续,$(0,\eta)$可导,\\
  $\therefore$由罗尔中值定理，存在一点$\xi \in （0,\eta）$,使\\
  $F^{\prime}(\xi)=2\xi{e}^{-\xi^{2}}{f}(\xi)+e^{-\xi^{2}}f^{\prime}(\xi)=0$\\
  $\therefore$$f^{\prime}(\xi)=2\xi f(\xi)$
\end{solution}
    
\begin{problem}[本题15分]
    设$f(x)$在$\left(-\infty,+\infty\right)$上连续,且$\displaystyle\lim_{x\to+\infty}f(x)=A,\displaystyle\lim_{x\to-\infty}f(x)=B$,($A,B$均为有限数)\\
    证明：\\
    1)$f(x)$在$\left(-\infty,+\infty\right)$上有界\\
    2)$f(x)$在$\left(-\infty,+\infty\right)$上一致连续
    \end{problem}
\begin{solution}[本题15分]\\
    1)$\because$$\displaystyle\lim_{x\to+\infty}f(x)=A,\displaystyle\lim_{x\to-\infty}f(x)=B$\\
    $\therefore$对$\varepsilon=1,\exists N>0$,当$x>N$时,$|f(x)-A|<1$\\
    $\therefore$$A-1<f(x)<A+1$\\
    对$\varepsilon=1,\exists M>0$,当$x<-M$时,$|f(x)-B|<1$\\
    $\therefore$$B-1<f(x)<B+1$\\
    又$\because$$f(x)$在$\left(-\infty,+\infty\right)$上连续,\\
    $\therefore$$f(x)$在$[-M,N]$上连续,\\
    $\therefore$$f(x)$在$[-M,N]$上有界\\
    $\therefore$$f(x)$在$\left(-\infty,+\infty\right)=(-\infty,-M)\cup[-M,N]\cup[N,-\infty)$上有界\\
    2)$\because$ $\displaystyle\lim_{x\to+\infty}f(x)=A$\\
    $\therefore$由柯西准则,$\forall\varepsilon>0,\exists N>0,$当$x_1,x_2>N$时,$|f(x_1)-f(x_2)|<\varepsilon$\\
    同理,$\forall\varepsilon>0,\exists M>0,$当$x_1,x_2<-M$时,$|f(x_1)-f(x_2)|<\varepsilon$\\
    又$\because$ $f(x)$在$[-M,N]$上连续,\\
    $\therefore$ $f(x)$在$[-M,N]$上一致连续,\\
    $\forall\varepsilon>0,\exists \delta>0,$当$x_1,x_2\in [-M,N]$时,$|x_1-x_2|<\delta$,$|f(x_1)-f(x_2)|<\varepsilon$\\
    $\therefore$在$\left(-\infty,+\infty\right)$上均有$|f(x_1)-f(x_2)|<\varepsilon$\\
    $\therefore$ $f(x)$在$\left(-\infty,+\infty\right)$上一致连续
\end{solution}
    

\begin{problem}[本题15分]
    设$f(x)$在$\left(-\infty,+\infty\right)$上二阶连续可导,且$f(0)=0$,若$g(x)=\left\{\begin{matrix}\displaystyle \frac{f(x)}{x},x\neq0\\ f^{\prime}(0),x=0\end{matrix}\right.$\\
    证明：\\
    1)$g(x)$在$\left(-\infty,+\infty\right)$上连续;\\
    2)$g(x)$在$\left(-\infty,+\infty\right)$上可微;\\
    3)$g^{\prime}(x)$在$\left(-\infty,+\infty\right)$上连续\\
\end{problem}
\begin{solution}
    证明：\\
    1)$\because$$f(x)$在$\left(-\infty,+\infty\right)$上二阶连续,则一定连续\\
    $\therefore$当$x\neq0$时,$g(x)$显然连续\\
    $\displaystyle\lim_{x\to0}g(x)=\displaystyle\lim_{x\to0}\frac{f(x)}{x}=\displaystyle\lim_{x\to0}\frac{f^{\prime}(x)}{1}=f^{\prime}(0)=g(0)$\\
    $\therefore$$g(x)$在$\left(-\infty,+\infty\right)$上连续\\
    2)$g^{\prime}(0)=\displaystyle\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\displaystyle\lim_{x\to0}\frac{f(x)-x f^{\prime}(0)}{x^{2}}=\frac{1}{2}f^{\prime\prime}(0)$\\
    $\therefore$$g^{\prime}(x)=\left\{\begin{array}{l l}{{\frac{xf(x)-f(x)}{x^{2}}}}&{{,x\neq0}}\\ {{\frac{1}{2}f^{\prime\prime}(0)}}&{{,x=0}}\\ \end{array}\right.$\\
    $\because$$f(x)$在$\left(-\infty,+\infty\right)$上二阶连续可导,$f(x),f^{\prime}(x)$均连续\\
    $\therefore$当$x\neq0$时,$g^{\prime}(x)$连续\\
    $\displaystyle\lim_{x\to0}g^{\prime}(x)=\displaystyle\lim_{x\to0}\frac{x f^{\prime}(x)-f(x)}{x^{2}}=\lim_{x\to0}\frac{f^{\prime}(x)+x f^{\prime}(x)-f^{\prime}(x)}{2x}=\lim_{x\to0}\frac{f^{\prime\prime}(x)}{2}=\frac{1}{2}f^{\prime\prime}(0)=g^{\prime}(0)$\\
    $\therefore$ $g^{\prime}(x)$在$\left(-\infty,+\infty\right)$上连续
\end{solution}

\begin{problem}[本题15分]
    设$f(x)$在$[0,2]$上连续,在$(0,2)$上二阶可导,且$\displaystyle\lim_{x\to1}\frac{f(x)}{x-1}=0$,$\displaystyle\int_{1}^{2}f(x)d x=f(2)$,证明：$\exists\xi\in(0,2)$,使得$f^{{\prime}{\prime}}(\xi)=0$\\
\end{problem}
\begin{solution}[本题15分]
    证明:\\
    $\because$ $\displaystyle\lim_{x\to1}\frac{f(x)}{x-1}=0$\\
    $\therefore$可得$f(1)=0,f^{\prime}(1)=0$\\
    $\because$ $f(x)$连续\\
    $\therefore$ $\exists \eta_{1}\in(1,2)$,使得$\int_{1}^{2}f(x)d x=f(\eta_{1})=f(2)$\\
    $\therefore$由罗尔中值定理，存在一点$\eta_{2} \in (\eta_{1},2)$,使得$f^{\prime}(\eta_2)=0$\\
    又$\because$$f(x)$二阶可导\\
    $\therefore$由罗尔中值定理，存在一点$\xi \in (1,\eta_2)\subset (0,2)$,使得$f^{\prime}(\xi)=0$\\
\end{solution}


\begin{problem}[本题15分]
    设$f(x)$在$\left[1,+\infty\right)$上单调递增,且无穷积分$\displaystyle\int_{1}^{+\infty}f(x)d x$收敛，证明$\lim\limits_{x\to+\infty}x f(x)=0$\\
\end{problem}
\begin{solution}[本题15分]
    证明:\\
    $\because$$f(x)$在$\left[1,+\infty\right)$上单调递增,$\int_{1}^{+\infty}f(x)d x$收敛\\
    则一定有$f(x)\leq0$,否则$\exists x=b>1$,使$f(x)>0$,则当$x>b$时,$f(x)>f(b)>0$\\
    $\therefore$$\int_{1}^{+\infty}f(x) d x=\int_{1}^{b}f\left(x\right)d x+\int_{b}^{+\infty}f(x) d x$发散与收敛矛盾\\
    $\therefore$$f(x)\leq0$\\
    $\because$$\int_{1}^{+\infty}f(x)d x$收敛\\
    $\therefore$$\forall\xi>0,\exists M>1$,当$x_{1},x_{2}>M$时,取$x_{1}=\frac{x}{2},x_{2}=x$\\
    且 $\because$$f(x)$单增\\
    $\therefore$$\varepsilon>\int_{\frac{x}{2}}^{x}f(t)d t\ge\int_{\frac{x}{2}}^{x}f(\frac{x}{2})d t=\frac{x}{2}f(\frac{x}{2})$\\
    $\therefore$$\frac{x}{2}f(\frac{x}{2})<\varepsilon$\\
    $\therefore$ $\displaystyle\lim_{u\to+\infty}u f(u)=0$,即有$\lim\limits_{x\to+\infty}x f(x)=0$
\end{solution}


\begin{problem}[本题15分]
    设$a_{0}=3$,对任意$n\geq1$,$\displaystyle na_{n} =\frac{2}{3}a_{n-1}-(n-1)a_{n-1}$,证明:
    当$|x|<1$时，幂级数$\displaystyle\sum_{n=1}^{n}a_{n}x^{n}$收敛,并求其和函数.\\
\end{problem}
\begin{solution}[本题15分]
    $\because$$na_{n} =\frac{2}{3}a_{n-1}-(n-1)a_{n-1}=(\frac{5}{3}-n)a_{n-1}$\\
    $\therefore$$\frac{a_{n}}{a_{n-1}}=\frac{\frac{5}{3}-n}{n}$\\
    $\therefore$ $\rho=\displaystyle\lim_{n\rightarrow\infty}\Big|\frac{a_{n}}{a_{n+1}}\Big|=\displaystyle\lim_{n\rightarrow\infty}\Big|\frac{\frac{5}{3}-n}{n}\Big|=1$\\
    $\therefore$ $R=1$\\
    $\therefore$$\displaystyle\sum_{n=1}^{n}a_{n}x^{n}$在$(-1,1)$内收敛\\
    设$S(n)=\displaystyle\sum_{n=1}^{\infty}a_nx^{n}$,\\
    $\begin{aligned}
        &S^{\prime}(x)=\sum_{n=1}^{\infty}n a_{n}x^{n-1}\\
        &=\sum_{n=1}^{\infty}(\frac{2}{3}a_{n-1}-(n-1)a_{n-1})x^{n-1}\\
        &=\frac{2}{3}\sum_{n=1}^{\infty}a_{n-1}x^{n-1}-x\sum_{n=1}^{\infty}(n-1)a_{n-1}x^{n-2}\\
        &=\frac{2}{3}\left(3+\sum_{n=2}^{\infty}a_{n-1}x^{n-1}\right)-x(0+\sum_{n=2}^{\infty}(n-1)a_{n-1}x^{n-2})\\
        &=2+\frac{2}{3}S(x)-x S^{\prime}(x)
    \end{aligned}$\\
    $\therefore$$(1+x)S^{\prime}(x)=2+\frac{2}{3}S(x)$\\
\end{solution}


\begin{problem}[本题15分]
    设$f(x,y)=x y g(x,y)$,其中$g(x,y)$在$(0,0)$某邻域内连续且$g(0,0)=0$,证明$f(x,y)$在$(0,0)$处可微.\\
\end{problem}
\begin{solution}[本题15分]\\
    证明：\\
    $f_{x}(x,y)=\displaystyle\lim_{x\to0}\frac{f(x,0)-f(0,0)}{x}=0$\\
    $f_{y}(x,y)=\displaystyle\lim_{y\to0}\frac{f(0,y)-f(0,0)}{y}=0$\\
    $\begin{aligned}
    &\lim_{\overset{x\rightarrow0}{y\rightarrow0}}\frac{f(x,y)-f(0,0)-f_{x}(0,0)x-f_{y}(0,0)y}{\sqrt{x^{2}+y^{2}}}\\
    &= \lim_{\overset{x\rightarrow0}{y\rightarrow0}}\frac{xyg(x,y)}{\sqrt{x^{2}+y^{2}}}\\
    &=\lim_{\overset{x\rightarrow0}{y\rightarrow0}}\frac{xy}{\sqrt{x^{2}+y^{2}}}\lim_{\overset{x\rightarrow0}{y\rightarrow0}}g(x,y)\\
    &=g(0,0)\lim_{\overset{x\rightarrow0}{y\rightarrow0}}\frac{kx}{\sqrt{1+k^{2}}}\\
    &=0
    \end{aligned}$
\end{solution}
\end{document}